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In many situations of practical importance, heat is generated internally at uniform rate within the conducting medium itself.

**Notable examples are: **

(i) Resistance heating in electrical appliances; essentially it is the conversion of electrical energy into thermal energy in the current carrying medium

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(ii) Energy generated in the fuel element of a nuclear reactor

(iii) Liberation of energy due to some exothermic chemical reactions occurring within the medium

(iv) Drying and setting of concrete

The rate of heat generation has to be controlled one; otherwise the resulting temperature growth might result in the failure of the medium. Undoubtedly temperature distribution within the medium and the rate of heat dissipation to the surroundings assume great importance in the design of thermal units.

**Plane Wall with Uniform Heat Generation****: **

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Consider heat conduction through a plane wall in which heat sources are uniformly distributed over the entire volume. The wall surfaces are maintained at temperature t_{1} and t_{2}, and the wall thickness δ is small in comparison with other dimensions.

**Assumptions: **

1. Steady state conditions

2. One-dimensional heat flow

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3. Constant thermal conductivity k

4. Uniform volumetric heat generation (q_{g} per unit volume) within the wall.

The differential equation describing the temperature distribution can be set up by making an energy balance on an elemental strip of thickness dx at a distance x from the left hand face of the wall.

Q_{x} (heat conducted in at distance x) = -kA dt/dx

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Q_{g} (heat generated in the element) = A dx q_{g}

Q_{x+dx} (heat conducted out at a distance x + dx) = Q_{x} + d/d_{x} (Q_{x}) dx

From steady state condition of heat flow,

**A. Both Surfaces are maintained at a Common Temperature:**

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From the prescribed boundary conditions,

t = t_{1} = t_{w} at x = 0

and t = t_{2} = t_{w} at x = δ

the constants of integration take the values

C_{2} = t_{w}; C_{1} = q_{g}/2k δ

Substituting these values of C_{1} and C_{2} in equation 4.3, the expression for temperature profile becomes-

Maximum temperature and its location within the wall can be worked out by differentiating the equation for temperature profile with respect to x and equating the derivative to zero.

The temperature distribution as prescribed by equation 4.4. is thus parabolic and symmetrical about the mid plane. Maximum value of temperature occurs at x = δ/2 and it equals

Heat transmission then occurs towards both surfaces, and for each surface it is given by-

For both surfaces

Q = A δ q_{g}

= (volume of conducting medium) (heat generating capacity)

Heat conducted to the wall surface is finally dissipated to the surrounding atmosphere at temperature t_{a}. Then for each surface,

Substituting this value of wall temperature in equation 4.4, one gets the temperature distribution in term of temperature t_{a} of the surrounding atmosphere

Equation 4.8. applies equally well to plane walls which are perfectly insulated at one face and maintained at a fixed temperature t_{w} on the other face.

**The full hypothetical slab with thickness 2δ can be considered with the same bounding conditions: **

The location x = δ corresponds to mid-plane of the hypothetical full wall or insulated face of the given wall.

The equation 4.4. for temperature distribution and the equation 4.5. for maximum temperature at the mid-plane (insulated end of given wall) may be written as-

**B. Temperatures of both the Surfaces are Different: **

When the boundary conditions-

t = t_{1} at x = 0 and

t = t_{2} at x = δ

are applied to equation 4.3, the constants of integration take the values-

Substituting these values of integration constants in equation 4.3, the expression for temperature profile becomes-

**Let us now examine two cases: **

(i) Maximum temperature occurs within the wall; the heat flow will then be from both the surfaces and the total heat flow will become-

Q_{t} = Q_{1} + Q_{2}

(ii) Maximum temperature occurs at the left hand face, i.e., t_{1} is maximum; the heat flow will then be only towards the right, i.e., in the direction of falling temperature

Q_{1 }= Q_{2} (only)

= kA [t_{1} – t_{2}/δ + q_{g}/2k δ]

If there is no internal heat generation (i.e., q_{g} = 0), the above expression reduces to heat conduction equation Q = kA (t_{1} – t_{2})/δ for a plane wall without any internal generation of energy.

**C. Current Carrying Electrical Conductor: **

When heat generated within a material is due to passage of electric current, q_{g} can be expressed in electrical terms-

where I is the current, R is the electrical resistance, ρ is resistivity, L and A are the length and cross-sectional area of the conductors. Combining these relations, we get-

Here i is the current density and electrical conductivity k_{c} is the reciprocal of the resistivity ρ of the medium.

**Example 1**:

The rear window of an automobile is made of 5 cm thick glass of thermal conductivity 0.8 W/m-deg. To defrost this window, a thin transparent film type heating element has been fixed to its inner surface. For the conditions given below, determine the electric power that must be provided per unit area of window if a temperature 5°C is maintained at its outer surface.

Interior air temperature and the corresponding surface coefficient, = 20°C and 12 W/m^{2}-deg. Surrounding air temperature and the corresponding surface coefficient, = – 15°C and 70 W/m^{2}-deg.

Electric heater provides uniform heat flux.

**Solution: **

Given: t_{i} = 20°C; h_{i} = 12 W/m^{2}-deg; t_{0} = – 15°C; h_{0} = 70 W/m^{2}-deg; t_{s} = 5°C

For unit area, the heat balance provides

**Example 2: **

A composite slab consists of 5 cm thick layer of steel (k = 146 kJ/m – hr-deg) on the left side and a 6 cm thick layer of brass (k = 276 kJ/m-hr-deg) on the right hand side. The outer surfaces of the steel and brass layer are maintained at 100°C and 50°C respectively. The contact between the two slabs is perfect and heat is generated at the rate of 4.2 × 10^{5} kJ/m^{2}-hr at the plane of contact. The heat thus generated is dissipated from both sides of composite slab for steady state conditions. Calculate the temperature at the interface and heat flow through each slab.

**Solution: **

Let t_{i} be the temperature at the interface. Under stipulation for heat dissipation from both sides,

**Cylinder with Uniform Heat Generation****: **

Consider heat conduction through a long and cylindrical rod of radius R and length L.

**Assumptions:**

1. Steady state conditions

2. One-dimensional radial conduction

3. Constant thermal conductivity k

4. Uniform volumetric heat generation per unit volume) within the solid.

The situation corresponds to a current carrying wire or a fuel element in a nuclear reactor. The appropriate differential equation describing the temperature distribution can be obtained by making an energy balance on a cylindrical shell of thickness dr and located at radius r.

Equation 4.12 may also be obtained from equation 2.25 by assuming steady state unidirectional heat flow in the radial direction.

Upon integration-

Another integration gives the general solution for temperature distribution-

**The constants of integration are to be determined from the relevant boundary conditions which are: **

(i) t = t_{w} at r = R

(ii) Heat generated equals the heat lost by conduction at the surface

Another condition stems from the fact that for a solid cylinder the centre line is the line of symmetry for the temperature distribution and there the temperature gradient must be zero. This condition of dt/dr = 0 at r = 0 is satisfied automatically when the two boundary conditions are satisfied.

The temperature gradient at the surface is-

With these values of integration constants, the general solution for the temperature distribution becomes-

Undoubtedly the temperature distribution is parabolic and the maximum temperature t_{max }occurring at the centre (r = 0) of the rod is given by-

Combining equation 4.18 and 4.19, we obtain the temperature distribution in the dimensionless form.

From overall energy balance, the rate at which energy is generated within the solid cylinder must be balanced by the rate at which energy leaves by convection at the cylinder boundary,

Substituting this value of surface temperature t_{w} in equation 4.18, one gets the temperature distribution in terms of temperature t_{a} of the surrounding atmosphere,

**Example 3: **

A 25 mm diameter meat roll (k = 1 W/m-deg) is roasted with the help of microwave heating. For good quality roasting, it is desired that temperature at the centre of roll is maintained at 100°C when the surrounding temperature is 25°C. What should be the heating capacity in W/m^{3} of the microwave if the heat transfer coefficient on the surface of meat roll is 20 W/m^{2}-deg? Also calculate surface temperature of the roll.

**Solution: **

Maximum temperature occurs at the centre of the roll and it is prescribed by the relation,

(b) The rate at which energy is generated within the cylindrical roll is balanced by the rate at which energy leaves by convection at the roll boundary. That is

**Sphere with Uniform Heat Generation****: **

Consider heat conduction through a solid sphere of radius R.

**Assumptions: **

1. Steady state conditions

2. One-dimensional radial conduction

3. Constant thermal conductivity k

4. Uniform volumetric heat generation (q_{g} per unit volume) within the solid.

The differential equation describing the temperature distribution can be set up by making energy balance on a spherical shell of thickness dr, at radius r-

Another integration gives the general solution for the steady-state radial temperature distribution,

Corresponding to centre of the sphere r = 0 and that gives C_{2} = 0. Applying the boundary condition t = t_{w} at r = R to expression (ii) and noting that C_{2} = 0, we get-

With these values of integration constants, the general solution for temperature distribution becomes-

Undoubtedly the temperature distribution is parabolic and the maximum temperature occuring at the centre (r = 0) of the sphere is given by-

Combining expressions (iii) and (iv), we obtain the temperature distribution in dimensionless form-

Undoubtedly, the heat conducted is equal to heat generated. But for steady state condition, the heat conducted (or generated) must be equal to heat convected from the outer surface of the sphere.

Substituting this value of surface temperature t_{w} in expression (iii), one gets the temperature distribution in terms of temperature t_{a} of the surrounding atmosphere

**Example 4: **

An 8 cm diameter orange, approximately spherical in shape, undergoes ripening process and generates 18000 kJ/m^{3}-hr of energy. If external surface of the orange is at 6.5°C, make calculations for temperature at the centre of the orange. Also determine the heat flow from the outer surface of the orange. Take thermal conductivity k = 0.8 kJ/m-hr-deg for the orange material.

**Solution:**